Answer:
(a) 
(b) 
(c) 
Step-by-step explanation:
We have given velocity as function of t 
Acceleration is equation rate if change of velocity with respect to time
So 
(a) Acceleration at t = 5 sec
(b) Acceleration at t = 10 sec
(c) Acceleration at t = 20 sec
25. he goes up by one foot (3 minus 2) each day and night, so 25.
you answer is 10 hope this help[s :D
Answer:
average rate of change is 3/2
Step-by-step explanation:
Recall that the average rate of change in an interval [x1, x2], is defined as:
[f(x2)-f(x1)] / (x2 - x1)
which for the interval [0, 2] becomes:
[f(2)-f(0)] / (2 - 0)
From the graph we read that f(0)= 1 and f(2) = 4, therefore:
[f(2)-f(0)] / (2 - 0) = [4 - 1] / 2 = 3/2
You can either use the inverse function theorem or compute the general derivative using implicit differentiation. The first method is slightly faster.
The IFT goes like this: if f(x) is invertible and f(a) = b, then finv(b) = a (where "finv" means "inverse of f").
By definition of inverse functions, we have
f(finv(x)) = finv(f(x)) = x
Differentiating both sides of the second equality with respect to x using the chain rule gives
finv'(f(x)) * f'(x) = 1
When x = a, we get
finv'(b) * f'(a) = 1
or
finv'(b) = 1/f'(a)
Now let f(x) = sin(x), which is invertible over the interval -π/2 ≤ x ≤ π/2. In the interval, we have sin(x) = √3/2 when x = π/3. We also have f'(x) = cos(x).
So we take a = π/3 and b = √3/2. Then
arcsin'(√3/2) = 1/cos(π/3) = 1/(1/2) = 2
